# Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 359: 8

54 cm

#### Work Step by Step

The diagram consist of two right triangles ADB and BDC By pythagoras theorem on ADB $AD^{2}$ =$DB^{2}$ + $AB^{2}$ $20^{2}$ = $DB^{2}$ + $16^{2}$ 400 = $DB^{2}$+ 256 $DB^{2}$ = 400 - 256 $DB^{2}$ = 144 DB = 12 By pythagoras theorem on DBC $CB^{2}$ =$DB^{2}$ + $DC^{2}$ $CB^{2}$ = $12^{2}$ + $5^{2}$ $CB^{2}$ = 144 + 25 $CB^{2}$ = 169 CB = 13 The perimeter of ABCD = AB + BC+CD+DA = 20 +5 + 13 + 16 =54 cm

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